Math 523H #3

# Math 523H #3 - Math 523H Homework#3 Sections 1.4 2.1 1.4#1...

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Math 523H Homework #3 Sections 1.4, 2.1 1.4 #1. Give 5 examples which show that P implies Q does not necessarily mean that Q implies P. If x R and x > 0, then x 2 > 0. However, if x 2 > 0, x is not necessarily greater than 0. If x R and x 0, then | x | ≥ 0. But if | x | ≥ 0, then x is not necessarily greater than or equal to 0. If x and y are even integers, then x + y is an even integer. But if x + y is an even integer, then x and y are not necessarily even. If a and b are rational numbers, then a b is a rational number. But if a b is a rational number, then a and b are not necessarily rational. If x and y are both even integers, then xy is an even integer. But if xy is an even integer, then it is not necessarily true that both x and y are even. 1.4 #4. Suppose that 0 < a < b . Prove that: (a) a < ab < b . Proof. a < b . Thus, a · a < a · b , or a 2 < ab . Since a > 0 and b > 0, ab > 0 so ab exists in R . Thus, we can are able to take the square root of both sides, and we get a 2 < ab , or a < ab . Since a < b , we also have a · b < b · b , or ab < b 2 . So ab < b 2 , or ab < b . By transitivity, a < ab < b . (b) ab 1 2 ( a + b ). Proof. For any a, b R , 0 ( a - b ) 2 , or 0 a 2 - 2 ab + b 2 . By (O4) of section 1.1, 0 + 4 ab ( a 2 - 2 ab + b 2 ) + 4 ab , or 4 ab a 2 + 2 ab + b 2 . But a 2 + 2 ab + b 2 = ( a + b ) 2 , so 4 ab ( a + b ) 2 . Since a > 0 and b > 0, 4 ab > 0 and thus 4 ab exists in R . Thus, we can take the square root of both sides, and 4 ab p ( a + b ) 2 . This may be simplified to 2 ab < a + b . Since 1 2 > 0, we can use (O5) of section 1.1, and we get 1 2 (2 ab ) 1 2 ( a + b ), or ab 1 2 ( a + b ). 1.4 #7. Prove that for any n natural numbers, x 1 , x 2 , ..., x n , | x 1 + x 2 + ... + x n | ≤ | x 1 | + | x 2 | + ... + | x n | . Proof. For n = 1, | x 1 | = | x 1 | , so the inequality holds for n = 1. Assume the above statement is also true for the first n natural numbers. | x 1 + x 2 + ... + x n + x n +1 | = | ( x 1 + x 2 + ... + x n )+ x n +1 | , so by the Triangle Inequality, | ( x 1 + x 2 + ... + x n )+ x n +1 | ≤ | x 1 + x 2 + ... + x n | + | x n +1 | . Because we assumed that the statement above was true for the first n natural numbers, | x 1 + x 2 + ... + x n | ≤ | x 1 | + | x 2 | + ... + | x n | . Therefore, | x 1 + x 2 + ... + x n | + | x n +1 | ≤ | x 1 | + | x 2 | + ... + | x n | + | x n +1 | . By transitivity, | x 1 + x 2 + ... + x n + x n +1 | ≤ | x 1 | + | x 2 | + ... + | x n | + | x n +1 | , so the statement is also true for the first n + 1 natural numbers. By induction, we have proven that | x 1 + x 2 + ... + x n | ≤ | x 1 | + | x 2 | + ... + | x n | for any n N . 1

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2 1.4 #8. Prove that for all positive integers n , 1 3 + 2 3 + ... + n 3 = (1 + 2 + 3 + ... + n ) 2 . Proof. For n =1, we have 1 3 = 1 = 1 2 . So the statement holds for n = 1. Now assume the above statement is true for the first n natural numbers, that is 1 3 + 2 3 + ... + n 3 = (1 + 2 + 3 + ... + n ) 2 . By (O4), 1 3 + 2 3 + ... + n 3 + ( n + 1) 3 = (1 + 2 + 3 + ... + n ) 2 + ( n + 1) 3 . But by Proposition 1.4.3, 1 + 2 + ... + n = n ( n +1) 2 , so we have 1 3 + ... + n + ( n + 1) 3 = ( n ( n +1) 2 ) 2 + ( n + 1) 3 = n 2 ( n +1) 2 4 + 4( n +1) 3 4 = ( n +1) 2 ( n 2 +4 n +4) 4 = ( n +1) 2 ( n +2) 2 4 = ( ( n +1)( n +2) 2 ) 2 = (1+2+ ... + n +( n +1)) 2 . Therefore, 1 3 + ... + n 3 +( n +1) 3 = (1+2+ ... + n +( n +1)) 2 , so the statement is also true for the first n +1 natural numbers. By induction, we have proven that 1 3 +2 3 + ... + n 3 = (1+2+3+ ... + n ) 2 .
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